 # IIIIIIIV C. Johannesson The Mole I. Molar Conversions.

## Presentation on theme: "IIIIIIIV C. Johannesson The Mole I. Molar Conversions."— Presentation transcript:

IIIIIIIV C. Johannesson The Mole I. Molar Conversions

C. Johannesson A. What is the Mole? n A counting number (like a dozen) n Avogadro’s number (N A ) n 1 mol = 6.02  10 23 items A large amount!!!!

C. Johannesson n 1 mole of hockey pucks would equal the mass of the moon! A. What is the Mole? n 1 mole of pennies would cover the Earth 1/4 mile deep! n 1 mole of basketballs would fill a bag the size of the earth!

C. Johannesson B. Molar Mass n Mass of 1 mole of an element or compound. n Atomic mass tells the...  atomic mass units per atom (amu)  grams per mole (g/mol) n Round to 2 decimal places

C. Johannesson B. Molar Mass Examples n carbon n aluminum n zinc 12.01 g/mol 26.98 g/mol 65.39 g/mol

C. Johannesson B. Molar Mass Examples n water n sodium chloride H2OH2O  2(1.01) + 16.00 = 18.02 g/mol  NaCl  22.99 + 35.45 = 58.44 g/mol

C. Johannesson B. Molar Mass Examples n sodium bicarbonate n sucrose  NaHCO 3  22.99 + 1.01 + 12.01 + 3(16.00) = 84.01 g/mol  C 12 H 22 O 11  12(12.01) + 22(1.01) + 11(16.00) = 342.34 g/mol

C. Johannesson C. Molar Conversions molar mass (g/mol) MASS IN GRAMS MOLES NUMBER OF PARTICLES 6.02  10 23 (particles/mol)

C. Johannesson C. Molar Conversion Examples n How many moles of carbon are in 26 g of carbon? 26 g C 1 mol C 12.01 g C = 2.2 mol C

C. Johannesson C. Molar Conversion Examples n How many molecules are in 2.50 moles of C 12 H 22 O 11 ? 2.50 mol 6.02  10 23 molecules 1 mol = 1.51  10 24 molecules C 12 H 22 O 11

C. Johannesson C. Molar Conversion Examples n Find the mass of 2.1  10 24 molecules of NaHCO 3. 2.1  10 24 molecules 1 mol 6.02  10 23 molecules = 290 g NaHCO 3 84.01 g 1 mol

IIIIIIIV C. Johannesson II. Molarity The Mole

C. Johannesson A. Molarity n Concentration of a solution. total combined volume substance being dissolved

C. Johannesson A. Molarity 2M HCl What does this mean?

C. Johannesson B. Molarity Calculations molar mass (g/mol) 6.02  10 23 (particles/mol) MASS IN GRAMS MOLES NUMBER OF PARTICLES LITERS OF SOLUTION Molarity (mol/L)

C. Johannesson B. Molarity Calculations n How many grams of NaCl are required to make 0.500L of 0.25M NaCl? 0.500 L0.25 mol 1 L = 7.3 g NaCl 58.44 g 1 mol

C. Johannesson B. Molarity Calculations n Find the molarity of a 250 mL solution containing 10.0 g of NaF. 10.0 g 1 mol 41.99 g = 0.238 mol NaF 0.238 mol 0.25 L M == 0.95M NaF

IIIIIIIV C. Johannesson III. Formula Calculations The Mole

C. Johannesson A. Percentage Composition n the percentage by mass of each element in a compound

C. Johannesson  100 = A. Percentage Composition %Cu = 127.10 g Cu 159.17 g Cu 2 S  100 = %S = 32.07 g S 159.17 g Cu 2 S 79.852% Cu 20.15% S n Find the % composition of Cu 2 S.

C. Johannesson %Fe = 28 g 36 g  100 = 78% Fe %O = 8.0 g 36 g  100 = 22% O n Find the percentage composition of a sample that is 28 g Fe and 8.0 g O. A. Percentage Composition

C. Johannesson n How many grams of copper are in a 38.0-gram sample of Cu 2 S? (38.0 g Cu 2 S)(0.79852) = 30.3 g Cu Cu 2 S is 79.852% Cu A. Percentage Composition

C. Johannesson  100 = %H 2 O = 36.04 g 147.02 g 24.51% H 2 O n Find the mass percentage of water in calcium chloride dihydrate, CaCl 2 2H 2 O? A. Percentage Composition

C. Johannesson B. Empirical Formula C2H6C2H6 CH 3 reduce subscripts n Smallest whole number ratio of atoms in a compound

C. Johannesson B. Empirical Formula 1. Find mass (or %) of each element. 2. Find moles of each element. 3. Divide moles by the smallest # to find subscripts. 4. When necessary, multiply subscripts by 2, 3, or 4 to get whole #’s.

C. Johannesson B. Empirical Formula n Find the empirical formula for a sample of 25.9% N and 74.1% O. 25.9 g 1 mol 14.01 g = 1.85 mol N 74.1 g 1 mol 16.00 g = 4.63 mol O 1.85 mol = 1 N = 2.5 O

C. Johannesson B. Empirical Formula N 1 O 2.5 Need to make the subscripts whole numbers  multiply by 2 N2O5N2O5

C. Johannesson C. Molecular Formula n “True Formula” - the actual number of atoms in a compound CH 3 C2H6C2H6 empirical formula molecular formula ?

C. Johannesson C. Molecular Formula 1. Find the empirical formula. 2. Find the empirical formula mass. 3. Divide the molecular mass by the empirical mass. 4. Multiply each subscript by the answer from step 3.

C. Johannesson C. Molecular Formula n The empirical formula for ethylene is CH 2. Find the molecular formula if the molecular mass is 28.1 g/mol? 28.1 g/mol 14.03 g/mol = 2.00 empirical mass = 14.03 g/mol (CH 2 ) 2  C 2 H 4